1. ## dual map dimension

F: field
V,W: F vector spaces
V*,W*: dual spaces (Hom(V,F);Hom(W,F))
φ : V -> W linear map
φ *:W*->V*
(λ: W -> F) |-> (λ∘ φ: V -> F)

Proof that dim(im(φ))=dim(im(φ*))
I have actually no idea how to proof it. I tried to use the rank-nullity theorem.
We know:
dim(V*)=dim(V)*dim(F)=dim(V)
dim(W*)=dim(W)*dim(F)=dim(W)

dim(V)=dim(ker(φ))+dim(im(φ))
dim(W)=dim(ker(φ*))+dim(im(φ*))
(r-n t)

dim(im(φ))<=dim(W)
dim(im(φ*))<=dim(V)

But how can I use these to show that the images have the same dimension? Maybe it is not possible how I tried it.

Is it somehow possible to show that there is an isomorphism between im(φ) and im(φ*)?

Has it something to do with the base of a dual space?

2. Hi,
you can't just use these formulas, you need more.

For instance, you can start by looking at $\ker \phi^*$. It consists of all $\lambda\in W^*$ such that $\lambda\circ\phi=0$. This condition is equivalent to ${\rm Im} \phi\subset \ker \lambda$: $\lambda$ has to be zero on ${\rm Im} \phi$, and may be equal to anything on a supplementary space. Thus the dimension of the set of such $\lambda\in W^*$ is $\dim W - \dim {\rm Im}\phi$. But we also have $\dim \ker \phi^* = \dim W - \dim {\rm Im }\phi^*$, hence we conclude $\dim {\rm Im} \phi=\dim {\rm Im} \phi^*$.

The statement about " $\lambda$ has to be zero on ${\rm Im} \phi$, and may be equal to anything on a supplementary space" and the computation of the dimension are made rigorous using bases or dual bases. For instance you can take a base of $W$ that starts with a base of ${\rm Im} \phi$, so that $\lambda\in\ker \phi^*$ iff it equals 0 on the first vectors of the basis. From there you can deduce that $\ker\phi^*$ is spanned by the elements of the dual base that correspond to vectors not in $\ker\phi^*$. (I know this is not very clear, that's just the rough idea, I let you try to find by yourself)